# Derivatives involving both product/quotient rule

• December 6th 2009, 09:33 AM
Archduke01
Derivatives involving both product/quotient rule
Find f'(x) for y = (x+5 / x+1) (2x+1)

I don't know how to proceed here. Do I use the quotient rule inside the first brackets? Then what?
• December 6th 2009, 09:43 AM
lvleph
First thing you can do is use the product rule
$y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$
.
From here we can take the derivative using the quotient rule on $\frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$.
Combine this with the previous result
$y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$
.
Now simplify.
• December 6th 2009, 09:51 AM
Archduke01
Quote:

Originally Posted by lvleph
First thing you can do is use the product rule
$y' = \frac{d}{dx}\left[\frac{x+5}{x+1}\right] \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$
.
From here we can take the derivative using the quotient rule on $\frac{d}{dx}\left[\frac{x+5}{x+1}\right] = \frac{(x+1) - (x + 5)}{(x+1)^2}$.
Combine this with the previous result
$y' = \frac{(x+1) - (x + 5)}{(x+1)^2} \cdot (2x+1) + 2\cdot
\frac{x+5}{x+1}$
.
Now simplify.

You don't end up with $2(x^2 + 2x + 3) / (x+1)^2$ do you?
• December 6th 2009, 10:10 AM
lvleph
Yes, that is what www.wolframalpha.com gives.